Calculate Speed of Rock after Falling 5 Seconds in Vacuum

In summary, we discussed the formulas for calculating force and velocity in different scenarios. For the hip sled, the formula for force necessary to lift the weight is F = M*g*sin(A), with A being the angle between the inclined plane and the ground. The formula for the falling rock on Earth is v = at, with a being the acceleration due to gravity (9.8 m/s^2). After 5 seconds, the rock would be going 49 m/s or 110 MPH.
  • #1
Tregg Smith
43
0
I used to know this stuff. How fast will a rock be going after falling 5 seconds in a vacuum?
I'd also like to know the formula in simple terms.

At the gym on the hip sled we call it- you lay back and push up with feet and legs. The rack supporting the weights is at 45 degrees. If you have 500# on it how much force is required to move it? I'd also like the formula for this as well as the answer. I think it's a simple 50% or 250# because it's at 45 degrees but not sure. I just don't like these kids bragging to an old guy that they are pushing 500#!

hip sled here:

AC038202l.jpg


Thank you! Tregg
 
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  • #2
if its 45 degrees. what is sin(45)? its more than 50%. Anyways, the answer is Fgsin(45).
 
  • #3
If the mass of the "weight" is M, gravitational acceleration is g, and the angle between the inclined plane and the ground is A, then the force necessary to lift the weight is

F = M*g*sin(A)

This corresponds to lifting a mass M' = F/g = M*sin(A)

Example:
M=100kg, g=10m/s^2, A=45 degrees gives M' = 100*sin(45) kg

Now, sin(45) = sqrt(2)/2, which is approx 0.7, so M' = 70 kg in this case.

If the angle A was 30 degrees, it would be half, since sin(30) = 1/2.

Torquil
 
  • #4
As for your question on the falling rock, I'm going to assume you mean on the surface of Earth, where gravity accelerates the rock at 9.8 m/s^2

thus for 5 seconds we have:

[tex] v = at = (9.8 \frac{m}{s^2}) (5 s) = 49 \frac{m}{s} = 110 MPH [/tex]
 
  • #5


I would be happy to provide a response to your questions.

To calculate the speed of a falling rock after 5 seconds in a vacuum, we can use the formula: v = gt, where v is the final velocity, g is the acceleration due to gravity (which is approximately 9.8 m/s^2), and t is the time in seconds. Plugging in the values, we get v = (9.8 m/s^2)(5 s) = 49 m/s. So after 5 seconds, the rock will be going 49 meters per second.

For the hip sled question, we can use the formula: F = mg sinθ, where F is the force required to move the sled, m is the mass of the sled (in this case, the weight on the sled), g is the acceleration due to gravity, and θ is the angle of the sled (45 degrees in this case). Plugging in the values, we get F = (500 lbs)(0.4536 kg/lb)(9.8 m/s^2) sin(45) = 2140 N. So the force required to move the sled with 500 lbs on it is approximately 2140 Newtons.

I hope this answers your questions and gives you the formulas you were looking for. Remember, it's not about bragging to others, but about understanding the science behind it. Keep learning and stay curious!
 

1. How do you calculate the speed of a rock after falling for 5 seconds in a vacuum?

To calculate the speed of a falling object, you can use the formula v = gt, where v is the velocity, g is the acceleration due to gravity (9.8 m/s^2), and t is the time the object has been falling. In this case, the time is 5 seconds. Therefore, the speed of the rock after falling for 5 seconds in a vacuum would be 49 m/s.

2. Why is it important to calculate the speed of a falling object?

Calculating the speed of a falling object is important for several reasons. It allows us to understand and analyze the motion of the object, which can be useful for predicting where it will land or how much force it will exert upon impact. It also helps us understand the effects of gravity and other forces on the object.

3. How does the vacuum affect the speed of a falling object?

In a vacuum, there is no air resistance or friction to slow down the object's fall. This means that the object will continue to accelerate at a constant rate due to gravity, resulting in a higher speed compared to falling in a non-vacuum environment.

4. Can the speed of a falling object in a vacuum be greater than the speed of light?

No, it is not possible for any object to exceed the speed of light, which is approximately 299,792,458 m/s. In a vacuum, the speed of a falling object will continue to increase, but it will never reach the speed of light due to the limitations of the laws of physics.

5. How can we use the calculated speed of a falling object in a vacuum in real-world applications?

The calculated speed of a falling object in a vacuum can be used in various real-world applications such as designing parachutes for skydiving or calculating the impact force of a falling object on a surface. It can also help in understanding the motion of objects in space where there is no air resistance or friction.

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